If $x, y \in (0, 1)$ and $x+y=1$, prove that $x\log(x)+y\log(y) \geq \frac {\log(x)+\log(y)} {2}$.
I transformed the LHS to $\log(x^xy^y)$ and the RHS to $\log(\sqrt{xy})$, from where we get that $x^xy^y \ge \sqrt{xy}$ beacuse the logarithm is a monotonically increasing function. From there we can transfrom the inequality into $x^{x-y}y^{y-x} \ge 1$. So here I am stuck.
I could have started from some known inequalities too, like the inequalities between means.
On
This can also be viewed as a special case of Chebyshev's sum inequality.
For any increasing function $f:I \to \Bbb R$ and $x, y \in I$ we have $$ \frac 12 \bigl( x f(x) + y f(y) \bigr) \ge \left( \frac 12(x+y) \right) \cdot \left( \frac 12(f(x)+f(y)) \right) \\ \implies x f(x) + y f(y) \ge \frac{x+y}{2} (f(x) + f(y)) \, . $$
More generally, if $x_1, \ldots, x_n \in I$ then $$ \sum_{k=1}^n x_k f(x_k) \ge \frac{1}{n} \sum_{k=1}^n x_k \sum_{k=1}^n f(x_k) \, . $$
Since $x-y$ and $\log x-\log y$ have the same sign, we have $$ (x-y)(\log x-\log y)\ge 0 $$ or equivalently $$ x\log x+y\log y\ge y\log x+x\log y. $$ Hence it holds that $$ 2x\log x+2y\log y\ge (x+y)\log x+(x+y)\log y=\log x+\log y. $$ This proves $$ y\log y+x\log x\ge \frac{\log x+\log y}{2}. $$
Note: As @Martin R pointed out, the result can be generalized to $$ x+y=1\Longrightarrow\; xf(x)+yf(y)\ge \frac{f(x)+f(y)}{2} $$ for any increasing function $f:(0,1)\to\Bbb R$.